67th Romanian Mathematical Olympiad

a) The hypothesis yields $\Delta ABE\sim \Delta CBD$, whence $\frac{BE}{BD}=\frac{AB}{CB}$, that is $\frac{EB}{AB}=\frac{BD}{BC}$. Since angles $ABC$ and $EBD$ have the same complement, $\widehat{ABC}=\widehat{EBD}$, hence $\Delta EBD\sim \Delta ABC$ (S.A.S.).

b) From $\Delta EBD\sim \Delta ABC$ follows that $\widehat{ACB}=\widehat{BDE}$, whence $$m(\widehat{DGC})=360^{\circ }-(m(\widehat{CBD})+m(\widehat{BDE})+m(\widehat{BCG}))=360^{\circ }-(m(\widehat{CBD})+m(\widehat{ACB})+m(\widehat{BCG}))=90^{\circ }.$$

$[GF]$ is a median in the right triangle $GAE$, hence $GF=AE/2$. Since $[AB]$ opposes to an angle of $30^{\circ }$ in the right triangle $BAE$, it follows that $AB=\frac{1}{2}AE$, hence $[AB]=[FG]$. In the same way, $[BC]=[GH]$.

From $△FGH\cong △FBH$ (S.S.S.) follows that $m(\widehat{FGH})=m(\widehat{FBH})=90^{\circ }-m(\widehat{HBD})-m(\widehat{CBF})$.

From $m(\widehat{HBD})=m(\widehat{HDB})=30^{\circ }$ and $m(\widehat{CBF})=m(\widehat{ABF})-m(\widehat{ABC})=60^{\circ }-m(\widehat{ABC})$, we get $m(\widehat{FGH})=m(\widehat{ABC})$, so $△FGH\cong △ABC$ (S.A.S.).